Timeline for Cofree Lie Coalgebra
Current License: CC BY-SA 3.0
13 events
| when toggle format | what | by | license | comment | |
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| Apr 20, 2017 at 9:31 | history | bumped | CommunityBot | This question has answers that may be good or bad; the system has marked it active so that they can be reviewed. | |
| Mar 21, 2017 at 9:03 | answer | added | Ender Wiggins | timeline score: 1 | |
| Dec 2, 2014 at 8:23 | history | undeleted | Mark.Neuhaus | ||
| Nov 17, 2014 at 6:40 | history | deleted | Mark.Neuhaus | via Vote | |
| Nov 16, 2014 at 22:50 | comment | added | Mark.Neuhaus | Yes you are absolutely right, to be pedantic about the conilpotent part. Without that things are very different. | |
| Nov 16, 2014 at 22:48 | comment | added | Mark.Neuhaus | I see. Hmm, was hoping for something more explicit. | |
| Nov 16, 2014 at 20:51 | comment | added | Dan Petersen | Also, your cofree Lie coalgebras should really be cofree conilpotent Lie coalgebras, to be pedantic. | |
| Nov 16, 2014 at 20:50 | comment | added | Dan Petersen | The cofreeness is also clear in this picture. A tree as above looks like it describes an $n$-fold iterated comultiplication, so "apply" this tree to an element of $C$ to get an element of $A^{\otimes n}$ which is only well defined modulo the antisymmetry and Jacobi relations. | |
| Nov 16, 2014 at 20:49 | comment | added | Dan Petersen | I don't have time to leave a detailed answer but this is explained in the book "Algebraic operads" by Loday-Vallette. In the first part of the book they explain that the tensor algebra quotiented by nontrivial shuffle products is naturally dual to the space of formal Lie words. You can think of $Lie^c(A) = Lie^c(n) \otimes_{S_n} A^{\otimes n}$ where $Lie^c(n)$ is the $S_n$-module spanned by binary trees with a single root and $n$ leaves, modulo the antisymmetry and Jacobi relations. The coproduct is given by a sum over all ways of splitting trees in half by deleting an edge. | |
| Nov 16, 2014 at 18:33 | comment | added | Mark.Neuhaus | Yes... Sloppy work | |
| Nov 16, 2014 at 18:32 | history | edited | Mark.Neuhaus | CC BY-SA 3.0 |
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| Nov 16, 2014 at 18:11 | comment | added | darij grinberg | The sum should be on the RHS, not the LHS. And you need to assume $p$ and $q$ positive when you factor out the span (not the set!) of the shuffle subspaces. | |
| Nov 16, 2014 at 18:05 | history | asked | Mark.Neuhaus | CC BY-SA 3.0 |